STRUCTURE OF SCANDIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications . Naturally occurring scandium (Sc) is composed of one stable isotope Sc-45. Twenty-four radioisotopes have been characterized with the most stable being Sc-46 with a half-life of 83.8 days, Sc-47 with a half-life of 3.35 days, and Sc-48 with a half-life of 43.7 hours. All of the remaining radioactive isotopes have half-lives that are less than four hours, and the majority of these have half-lives that are less than two minutes, the least stable being Sc-39 with a half-life shorter than 300 nanoseconds. The half-lives for isotopes with mass numbers less than 39 is unknown. WHY Sc-45 WITH S=-7/2 IS A STABLE NUCLIDE After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. In my paper STRUCTURE OF Sc-45 AND Ti-45 I explained carefully that the stable structure of Sc-45 is due to the fact that the three extra neutrons make two bonds per neutron able to overcome the pp and nn repulsions. They also increase the binding energies of bonds for overcoming the pp and nn repulsions. For example the unstable Sc-43 with S=-7/2 has only one extra neutron with two bonds which cannot give enough energy to bonds for overcoming the pp and nn repulsions. Also the Sc-42 which breaks the symmetry with S = +5 has not any extra neutron for increasing the binding energies of bonds . ' ' STRUCTURE OF Sc-43, Sc-41 Sc-39 AND Sc-37 WITH S =-7/2 After a careful analysis I found that the structure of the above nuclides is based on the structure of Sc-43 with S=-7/2. In the following diagram of Sc-43 you see that the core is not the parallelepiped of Mg-24. Here the core consists of a parallelepiped with the five horizontal planes which exist from the -HP1 to the -HP5 giving S = -3 . Here you also see that the deuteron p18n18 of -HP3 gives S = -1 . Whereas the two vertical pn systems with strong vertical bonds like the p19n19 and n20p20 give S=0. Note that the p19n19 is not shown, because it exists in front of the n5 and p7. Also the n20p20 is not shown because it exists behind the p6 and n8. It is of interest to notice that the p19 with the p4 and also the n20 with the n4 form blank positions for receiving the n21(+1/2) of +HP2 and the p21(+1/2) of the same horizontal plane . Also the p20 and p10 form a blank position for receiving the extra n22(-1/2) of the -HP5. That is the above 7 nucleons are not shown. Under this condition the total spin is given by S = -3 -1 + 0 +1/2 +1/2 -1/2 = -7/2 . Therefore in the absence of two neutrons of opposite spins we get the structure of Sc-41 with S=-7/2. Similarly in the structures of Sc-39 and Sc-37 with S=-7/2 we have 4 and 6 absent neutrons of opposite spins respectively. DIAGRAM OF Sc-43 WITH S =-7/2 The structure consists of five horizontal planes like the -HP1, +HP2 -HP3 +HP4 and -HP5. Here the n21, p19 and n19 are not shown because they are in front of p4, n5 and p7 respectively. Also the p21, n20, p20 and n22 are not shown because they are behind the n4, p6, n8 and p10 respectively ' ' ' p15.....n10......p10' ' -HP5 n15.......p9.......n9 ' ' n14.......p8........n8 ' ' +HP4 p14......n7........ p7......n23 ' ' p13.......n6........p6........n18' ' -HP3 n13.......p5........n5.......p18 ' ' n12.......p4........n4........p17' ' +HP2 p12......n3........p3.......n17 ' ' p11......n2........p2.......n16' ' -HP1 n11.....p1.......n1........p16 ' ' NUCLEAR STRUCTURE OF Sc-47, Sc-49, Sc-51, Sc-53, Sc-55, Sc-57 AND Sc-59 WITH S = -7/2' After a careful analysis I found that the structure of the above nuclides is based on the structure of Sc-45 with S =-7/2. For example the Sc-59 with S =-7/2 has 14 extra neutrons of opposite spins which make single bonds leading to the decay. ' ' NUCLEAR STRUCTURE OF Sc-44, Sc-42, Sc-40 AND Sc-38 ' For the unstable structure of Sc-44 with S =+2 you can read my STRUCTURE OF Ti-44, Sc-44 AND Ca-44 . However for the structure of Sc-42 with S =+5 you can see the following diagram in which the core is not the parallelepiped of Mg -24. Here the core consists of five horizontal planes like the +Hp1, -HP2, +HP3, -HP4 and +HP5 giving S =+3 . Also the deuteron p18n18 of the +HP3 gives S=+1. Note that the nucleons p19, n19 and p21 are not shown because they are in front of n5, p3 and n1 respectively. Also the nucleons n20, p20 and n21 are not shown because they are behind the p6, n4, and p2 respectively. They make strong vertical bonds with a spin S =+1. Under this condition the total spin is given by S = +3 + 1 + 1 = +5 Following this structure we conclude that the structure of Sc-40 with S= -4 is based on the structure of Sc-42 with S=+5 because all nucleons change the spins. For example we have -HP1, +HP2 -HP3 +HP4 and -HP5 giving S = -5. Then, we get the structure of SC-40 with S =-4 because we have two absent neutrons of positive spins . That is S = -5 - 2(+1/2) = -4 . Also the structure of Sc-38 with S =-2 is based on the structure of Sc-40 with S= -4 because in the absence of 2 more neutrons of positive spins than those of Sc-40 we get the structure of Sc-38 with S =-2. That is S = -4 - 2(+1/2) = -2. '''NUCLEAR STRUCTURE OF Sc-46, Sc-48, Sc-50, Sc-52, Sc-54, Sc-56 AND Sc-60 ' After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of Sc-42 with S = +5. For example the Sc-60 with S =+3 has 4 extra neutrons of negative spins and 14 neutrons of opposite spins giving S=0. They form single horizontal bonds of weak binding energy leading to the decay. In this case the total spin is given by S = +5 + 4(-1/2) + 0 = +3 . '''DIAGRAM OF Sc-42 WITH S = +5 In this structure we have 5 horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4 and +HP5. Here the nucleons p19, n19 and p21 are not shown because they are in front of n5, p3 and n1 respectively. Also the nucleons n20, p20 and n21 are not shown because they are behind the p6, n4 and p2 respectively. ' ' ' p15.....n10......p10' ' +HP5 n15.......p9.......n9 ' ' n14.......p8........n8 ' ' -HP4 p14......n7........ p7 ' ' p13.......n6........p6........n18' ' +HP3 n13.......p5........n5.......p18 ' ' n12.......p4........n4........p17' ' -HP2 p12......n3........p3.......n17 ' ' p11......n2........p2.......n16' ' +HP1 n11.....p1.......n1......p16 ' Category:Fundamental physics concepts